Answer:
There will be 52.37 grams of CO2 produced.
Step-by-step explanation:
Step 1: Data given
Mass of citric acid (C6H8O7) = 100.00 grams
Molar mass of citric acid = 192.12 g/mol
Mass of baking sode (NaHCO3) = 100.00 grams
Molar mass of baking soda = 84.01 g/mol
Step 2: The balanced equation
3NaHCO3 + C6H8O7 → 3CO2 + 3H2O + Na3C6H5O7
Step 3: Calculate moles of NaHCO3
Moles NaHCO3 = Mass NaHCO3/ molar mass NaHCO3
Moles NaHCO3 = 100.00 grams / 84.01 g/mol
Moles NaHCO3 = 1.190 moles
Step 4: Calculate moles of citric acid
Moles C6H8O7 = 100.00 grams / 192.12 g/mol
Moles C6H8O7 = 0.5205 moles
Step 5: Determine the limiting reactant
NaHCO3 is the limiting reactant. It will completely be consumed (1.190 moles).
Citric acid is in excess. There will react 1.190/3 = 0.3967 moles of C6H8O7
There will remain 0.5205 - 0.3967 = 0.1238 moles
Step 6: Calculate moles of CO2
For 3 moles of NaHCO3 we need 1 mol of C6H8O7 to produce 3 moles of CO2, 3 moles of H2O and 1 mol of Na3C6H5O7
For 1.190 moles of NaHCO3, we'll have 1.190 moles of CO2
Step 7: Calculate Mass of CO2
Mass of CO2 = Moles of CO2 * molar mass CO2
Mass CO2 = 1.190 moles * 44.01 g/mol
Mass CO2 = 52.37 grams
There will be 52.37 grams of CO2 produced.